From pg. 97 of Categories for the Working Mathematician, the author starts out with something that I understand:

To assert that a category $C$ has all finite products and coproducts is to assert that products, terminal, and initial and coproducts exist,

but then asserts something that don't understand:

thus the functors $C \rightarrow \mathbf{1}$ and $\Delta : C \rightarrow C \times C$ have both left and right adjoints. Indeed, the left adjoints give initial object and coproduct, respectively, while the right adjoints give terminal object and product, respectively.

Why is this?

  • $\begingroup$ Look at the functor $G: C\to \mathbf{1}$ for instance. It having a right adjoint means : there exists $F: \mathbf{1}\to C$, and a natural bijection $Hom(A, Fe)\simeq Hom(GA, e)$ where $A$ is any object in $C$ and $e$ is the unique object in $\mathbf{1}$. But $Hom(GA,e) =\{id_e\}$ by definition of $\mathbf{1}$ so for all $C$-objects $A$, $Hom(A,Fe)$ has size $1$: that's precisely saying that $Fe$ is a terminal object. Similarly, a left adjoint for $G$ provides an initial object. $\endgroup$ – Max Jun 29 '17 at 13:57
  • $\begingroup$ Then you can prove, using the definition of adjoints and (co) products that adjoints for $\Delta$ provide (co)-products $\endgroup$ – Max Jun 29 '17 at 13:58

You can verify by direct computation; e.g. $L$ is left adjoint to $\Delta$ iff

$$\begin{align} \hom_{C} (L(X,Y), Z) &\cong \hom_{C \times C}((X,Y), \Delta Z) \\&\cong \hom_{C \times C}((X, Y), (Z, Z)) \\&\cong \hom_{C}(X, Z) \times \hom_{C}(Y, Z) \end{align}$$

but this is precisely the universal property of the binary coproduct:

$$ \hom_C(X \amalg Y, Z) \cong \hom_{C}(X, Z) \times \hom_{C}(Y, Z) $$

and thus $L$ is left adjoint to $\Delta$ if and only if (binary coproducts exist and) there is a natural isomorphism $L(X,Y) \cong X \amalg Y$.

  • $\begingroup$ So -- just to confirm my understanding -- making $L$ the functor in which $L(X,Y) = X \amalg Y$ would imply that $L$ is left adjoint to $\Delta$; is this the case? $\endgroup$ – user1770201 Jun 29 '17 at 14:12
  • 1
    $\begingroup$ @user1770201: That is correct. $\endgroup$ – user14972 Jun 29 '17 at 14:17

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